Answer: D

Small Hint:

Since $99 = 3 \cdot 33,$ the equation is $m^3 + n^3 + (-33)^3$ $- 3mn(-33) = 0$

Big Hint:

Use $x^3 + y^3 + z^3$ $- 3xyz = (x + y + z)$ $\cdot (x^2 + y^2 + z^2 - xy - yz - zx)$

Solution:

Writing $z = -33,$ the equation becomes $m^3 + n^3 + z^3 - 3mnz = 0,$ which factors as $$\begin{aligned} &(m + n - 33) \\ &\quad {}\cdot \scriptsize\left(m^2 + n^2 + 33^2 - mn + 33m + 33n\right) \\ &= 0. \end{aligned}$$

The second factor equals $\scriptsize\tfrac12\big[(m - n)^2 + (m + 33)^2 + (n + 33)^2\big],$ which is $0$ only at $(m, n) = (-33, -33);$ this satisfies $mn \ge 0.$

Otherwise $m + n = 33.$ With $mn \ge 0$ both are nonnegative, giving $(0, 33), (1, 32), \ldots, (33, 0),$ which is $34$ pairs. Together there are $35$ solutions.

Thus, the correct answer is D.